If $\mathcal{B}^*$ is a basis for $V^*$, then $V^*$ is finite dimensional. Let $V$ be a vector space over some field K. It might be that $\dim(V) = \infty$. Let $\mathcal{B}$ be a basis for $V$. Let $V^*$ be the dual space for $V$ and $\mathcal{B}^*$ had to show the following:
If $\mathcal{B}^*$ is a basis for $V^*$, then $V^*$ is finite dimensional. 

What I thought
I find this one very hard because everything appears about this appears so abstract. Maybe I should show that $\mathcal{B}^*$ is finite, I would even say that this has to be shown.
Please provide me a hint. I don't know where to start.
 A: Let $B = \{ e_\alpha \}_{\alpha \in I}$ and denote $B^{*}$ by $B^* = \{ \varphi^{\alpha} \}_{\alpha \in I}$. The relation between $B$ and $B^*$ is given by $\varphi^\alpha(e_\beta) = \delta^{\alpha}_{\beta}$. If $V$ is infinite dimensional, you can choose some sequence of distinct elements $\alpha_i \in I$ and define a linear functional $\varphi \colon V \rightarrow \mathbb{F}$ be requiring that $\varphi(e_{\alpha_i}) = 1$ for all $1 \leq i < \infty$ while $\varphi(e_\alpha) = 0$ for all $\alpha \in ( I \setminus \{ \alpha_1, \alpha_2, \ldots \} )$. This is possible because $B$ is a basis for $V$. But then $\varphi \notin \mathrm{span}(B^*)$ because any functional in $\mathrm{span}(B^*)$ vanishes on all but finitely many elements of $B$. Thus, if $B^*$ is a basis for $V^*$, $V$ must be finite dimensional.
A: Recall that each of the elements of $\def\B{\mathcal B}\B^*$ is a linear form that is zero one all elements of $\B$ except one (say$~b$), where it takes the value$~1$. It is the coordinate function for that basis vector$~b$: when applied to any linear combination of the basis vectors$~\B$, it returns the coefficient that was used for$~b$ (or in case $b$ was not used at all it returns $0$). By definition of a basis, every $v\in V$ can be written uniquely as a linear combination of elements of$~\B$, so this gives a well defined description of the value of our element of$~\B^*$ on any element of$~V$.
The set $\B^*$ is always linearly independent. To see this one argues that for any linear combination $L$ of elements of$~\B^*$ that produces the zero function on$~V$, every coefficient used in $L$ must be zero (definition of linear independence). Fixing an element$~\beta$ of$~\B^*$ for which we shall consider the coefficient in $L$, it ($\beta$) is the coordinate function associated to some $b\in\B$, as described above. One can conversely recover the coefficient of $\beta$ in a linear combination$~L$ by applying the value of$~L$ (a linear function on$~V$) to the vector$~b$. This is because every term in$~L$ not involving$~\beta$ is taking the coordinate of some basis vector of$~\B$ other than$~b$, and for the vector$~b$ such coordinates are$~0$; the only term of$~L$ that affects its value at$~b$ is the one involving$~\beta$. But we supposed the value of$~L$ is the zero function, so applying it to$~b$ certainly gives us $0$, and this is the coefficient of$~\beta$ in$~L$; we have proven our claim.
Up until here, everything holds whether $\B$ is finite or not. But finiteness is built into the definition of linear combination: a linear combination of vectors from a set $S$ is obtained by picking finitely many elements of $S$, multiplying each by some coefficient, and adding up the results. (This is why above I added the clause for "in case $b$ is not used at all".) It cannot be otherwise, since in adding infinitely many vectors is not defined in general (consider adding up all the monomials $X^i$ in the vector space of polynomials in $X$). Occasionally it is useful to consider, for an infinite set$~S$, formally infinite linear combinations by attaching coefficients to all elements of$~S$, but this only has a meaning if only finitely many of those coefficients are nonzero; the meaning is then the linear combination of those nozero terms. 
This comes to bear when we investigate whether $\B^*$ generates (spans) the dual space$~V^*$. Given a linear form $\alpha\in V^*$, we have seen that if $\alpha$ can be written as (the value of) a linear combination $L$ of elements of$~\B^*$, then the coefficient in$~L$ of $\beta\in\B^*$ associated to some$~b\in\B$ must be equal to the value$~\alpha(b)$. But this is problemetic in general if $\B$ is infinite: it could be that $\alpha(b)\neq0$ for infinitely many different$~b\in\B$, and in that case one gets an impossible requirement for the linear combination$~L$. Whenever $\B$ is infinite, there are always such $\alpha\in V^*$ that are problematic, since linear functions are determined by their values at the elements of a given basis (here $\B$), which values can be arbirtary: restriction to$~\B$ gives a bijection between $V^*$ and the set $\mathcal F(\B,\Bbb R)$ of all functions $\B\to\Bbb R$ (sometimes also written $\Bbb R^\B$); one could for instance choose $\alpha$ with $\alpha(b)=1\neq0$ for all $b\in\B$. On the other hand, for those $\alpha$ for which $\alpha(b)\neq0$ occurs only for finitely many different $b\in\B$, one does get a valid linear combination$~L$ of elements of$~\B^*$, and the value of$~L$ coincides with$~\alpha$, since both take the same values for each $b\in\B$. This shows in particular that if $\B$ is finite, then $\B^*$ does span $V^*$.
Together (and using the fact that all bases of $V$ have the same number of elements) this shows that $\B^*$ is a basis of $V^*$ if and only if $\B^*$ spans $V^*$ if and only if $\B$ is finite if and only if $V$ is finite dimensional.
